Conditional-on-design bias of the empirical local-linear estimator

Put $h=h_n$, $U_i=(x-X_i)\mathrm{/}h$, $S_j=(nh{)}^{-1}{\sum }_{i}K(U_i)U_i^j$ and $r_h(u)=m(x-hu)-m(x)+hu{m}^{\mathrm{\prime }}(x)$. Then $S_0\to f(x)$, $S_2\to f(x){\mu }_2$ by [F2]. Symmetry and the first-order density expansion give $\mathbb{E}{S}_1=-h{f}^{\mathrm{\prime }}(x){\mu }_2+o(h)$; the tail is controlled by $u^{2}K$ under the global alternative. Also $\mathrm{Var}(S_1)\le C(nh{)}^{-1}\int u^2{K}^2$, using localization or the global density bound. In the rapid-tail alternative its local part has this bound, and the far contribution to $\mathrm{Var}(S_1)$ is at most $C{h}^2{n}^{-1}{\int }_{\mathrm{\mid }z-x\mathrm{\mid }>\delta }f(z)\mathrm{\mid }z-x{\mathrm{\mid }}^{-4}dz$, hence no larger than $O((nh{)}^{-1})$. Since $n{h}^3\to \mathrm{\infty }$, $(S_1-\mathbb{E}{S}_1)\mathrm{/}h\to 0$ in probability by the variance bound, so $S_1=O_P(h)$.

Set $R_j=(nh{)}^{-1}{\sum }_{i}K(U_i)U_i^j{r}_h(U_i)$. Taylor's integral formula gives $\mathrm{\mid }{r}_h(u)\mathrm{\mid }\le C{h}^2{u}^2$ and $r_h(u)\mathrm{/}{h}^2\to \frac{1}{2}{m}^{\mathrm{\prime }\mathrm{\prime }}(x)u^2$. The truncation proof of [F2] applies to this varying integrand: on any fixed set where $\mathrm{\mid }u\mathrm{\mid }$ and $K(u)$ are bounded it converges uniformly, its average has variance $O_A((nh{)}^{-1})$, and the absolute tail expectation is at most $C{\int }_{\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}}{u}^{2}K$. Hence $R_0\mathrm{/}{h}^2\to \frac{1}{2}{m}^{\mathrm{\prime }\mathrm{\prime }}(x)f(x){\mu }_2$ in probability. Furthermore $\mathrm{\mid }{r}_h(u)\mathrm{\mid }\le Ch\mathrm{\mid }u\mathrm{\mid }$ and $r_h(u)\mathrm{/}(h\mathrm{\mid }u\mathrm{\mid })\to 0$ pointwise. Splitting the integral as before gives $\mathbb{E}\mathrm{\mid }{R}_1\mathrm{\mid }\mathrm{/}h\le \int K(u)\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }{r}_h(u)\mathrm{\mid }f(x-hu)\mathrm{/}hdu\to 0$. Markov's bound yields $R_1=o_P(h)$. Under the rapid-tail alternative [F5] gives $\int \mathrm{\mid }m\mathrm{\mid }f<\mathrm{\infty }$; first apply these arguments on $\mathrm{\mid }{X}_i-x\mathrm{\mid }\le \delta$. For the far part put ${ϵ}_h={\sup }_{\mathrm{\mid }u\mathrm{\mid }\ge \delta \mathrm{/}h}\mathrm{\mid }u{\mathrm{\mid }}^{3}K(u)\to 0$ and expand $r=m(z)-m(x)-m^{\mathrm{\prime }}(x)(z-x)$. Its expected absolute contributions to $R_0\mathrm{/}{h}^2$ and $R_1\mathrm{/}h$ are bounded by $C_{\delta }{ϵ}_h\int (1+\mathrm{\mid }m\mathrm{\mid })f\to 0$. The same tail estimate gives the stated expectation expansion for $S_1$. Thus all limits hold without global smoothness.

On a nonsingular design, [F1] and [F3] give $b_n-m(x)=(S_2{R}_0-S_1{R}_1)\mathrm{/}(S_0{S}_2-S_1^2)$, since the constant and slope terms cancel exactly in the intercept. Steps 1.1–2.1 give $S_1{R}_1=o_P(h^2)$, and the denominator tends to $f(x{)}^2{\mu }_2>0$. Division gives the claimed coefficient. The singular-design event has probability tending to zero by [F2]; even after scaling by $h^{-2}$ the probability of any nonzero fallback contribution is at most that event probability, so it does not change convergence in probability. This argument integrates the response law separately for each design and never interchanges an undefined unconditional signed expectation.